Problem: The $n^{\text{th}}$ derivative of $g$ at $x=0$ is given by $g^{(n)}(0)=\dfrac{2^{n-1}}{n!}$ for $n\ge 1$. What is the coefficient for the term containing $x^4$ in the Maclaurin series of $g$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{1}{3}$ (Choice B) B $\dfrac{1}{144}$ (Choice C) C $\dfrac{1}{18}$ (Choice D) D $\dfrac{1}{72}$
Solution: The fourth-degree term of the Taylor series centered at $~x=0~$ is $~{g}\,^{\prime\prime\prime\prime}\left( 0 \right)\frac{{{x}^{4}}}{4!}\,$, $\Big($ also written $ ~g\,^{(4)}(0)\frac{x^4}{4!}\Big)$. From the given information, $~{g}\,^{\prime\prime\prime\prime}\left( 0 \right)=\frac{2^3}{4!}=\frac{8}{24}=\frac{1}{3}\,$. Then the coefficient of $~{{x}^{4}}~$ is $~\frac{1}{3}\cdot \frac{1}{4!}=\frac{1}{72}\,$.